RBSE Maths Class 8 Chapter 14: Important Questions and Solutions

RBSE Maths Chapter 14 – Area Mathematics Class 8 Important questions and solutions are available here. The additional, exercise important questions and solutions of Chapter 14, available at BYJU’S, contain step by step explanations. It can be used as study material for facing the board exams. All these important questions are based on the new pattern prescribed by the RBSE. Students can also get the syllabus and textbooks on RBSE Class 8 solutions.

Chapter 14 of the RBSE Class 8 Maths will help the students to solve problems related to the area of a trapezium, area of a rhombus, area of a quadrilateral, area of a polygon, area of an irregular polygon.

RBSE Maths Chapter 14: Additional Questions and Solutions

Question 1: State whether the following statements are true/false.

A] Area of a parallelogram, whose base is 6.5 cm and height 4 cm is 26 cm2.

B] 1 meter = 100 square cm.

C] 1 hectare = 10000 square meter.

D] If the diagonals of a rhombus are 24 cm and 7 cm, then the area of the rhombus is 168 cm2.

Question 1: The length of two perpendicular sides of a trapezium are 10 cm and 16 cm. The perpendicular distance between them is 8 cm. Find the area of the trapezium.

Solution:

Area of trapezium = [1 / 2] * height * [sum of the parallel sides]

= [1 / 2] * 8 * [10 + 16]

= [1 / 2] * 8 * 26

= [1 / 2] * 208

= 104

The area of the trapezium is 104 square centimeters.

Question 2: The roof of the building is in a particular shape, as shown in the figure below. If all the dimensions are of equal length, then find the area of the whole design.

Solution:

Every shape in the figure is a trapezium.

Number of shapes = 4

Area of one shape = [1 / 2] * height * [sum of the parallel sides]

= 2 * ([1 / 2] * (6 + 8) * 4)

= 2 * ([½] * 14 * 4]

= 2 * ([½] * 56)

= 2 * 28

= 56 cm2

Area of the entire shape = 4 * Area of one shape = 4 * 56 = 224cm2.

Question 3: The area and height of trapezium are 34cm2 and 4 cm. One of its parallel sides is 10cm. Find the length of another parallel side.

Solution:

Let the length of the required parallel side be ‘x’ cm.

Area of trapezium = [1 / 2] * height * [sum of the parallel sides]

34 = [1 / 2] * 4 * [10 + x]

34 = 2 * [10 + x]

34 = 20 + 2x

34 – 20 = 2x

14 = 2x

14 / 2 = x

7 = x

Hence, the length of another parallel side is 7 cm.

Question 4: The top surface of a platform is In the shape of a regular octagon, as shown in the figure. Find the area of the octagonal surface.

Solution:

Required Area = Area of trapezium above + Area of rectangle + Area of lower trapezium

= [1 / 2] * (5 + 11) * 4 + 11 * 5 + [1 / 2] * [11 + 5] * 4

= [1 / 2] * (16) * 4 + 55 + [1 / 2] * [16] * 4

= 32 + 55 + 32

= 119m2

Question 5: Length between the opposite vertex of the rhombus-shaped plot is 12.5 m and 10.4 m. Find the total cost of making this plot as a flat surface if the cost of making a flat surface per square meter is Rs. 180.

Solution:

Area of rhombus-shaped plot = [1 / 2] * product of diagonals

= [1 / 2] * [12.5 * 10.4]

= [1 / 2] * 130

= 65m2

Cost of making the plane this plot as a flat surface = 65 x Rs. 180 = Rs 11,700.

Question 6: Find the area of the combined rhombus-shaped tiles as given in the figure.

Solution:

Area of rhombus-shaped tiles = [1 / 2] * product of diagonals

= 2 * ([1 / 2] * [4.5] * 9)

= 2 * ([1 / 2] * 40.5)

= 2 * [20.25]

= 40.5 cm2

Question 7: The field of the Kalyan is in the form of a quadrilateral. The diagonal of this field is 220 m and, the perpendiculars dropped on it from the remaining opposite vertices are 80 m and 130, respectively. Find the area of the field.

Solution:

Area of quadrilateral shaped form = [1 / 2] * diagonal * [sum of the length of perpendicular on the diagonal]

= [1 / 2] * 220 * [80 + 130]

= [1 / 2] * 220 * 210

= [1 / 2] * 46200

= 23100 m2

Question 8: Fill in the blanks

(i) Area of the rhombus is _____ product of the diagonals.

(ii) Both the diagonals of the scalene quadrilateral is always _____.

(iii) Area of the _____ quadrilateral can be found through the formula height * ([sum of parallel sides] / 2)

(iv) The quadrilateral whose unequal diagonals bisect each other is called _____.

Question 2: Each side of the hexagon ABCDEF has a side of length 5 cm as given in the figure. Riya and Riema find the area of the region by dividing it into two parts into two different manners. Compare the area in both cases.

RBSE Maths Chapter 14: Additional Questions and Solutions

Question 1: Azhar has a trapezium-shaped farm. He divides it into three parts as shown in the figure below. Show that the area of the trapezium PQRS = area of triangle SPT + area of rectangle STUR + area of triangle RUQ. Also, compare it with the area of trapezium PQRS.